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Graphic Equalizer Series 
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Part 4

The goal of our RL simulator is to create an impedance to ground such that the current waveform looks like the RL current waveform.
RL SIMULATOR
Although the RC's current is very different from RL's current, the
voltage across C appears identical to the RL current! What if we could take
the voltage across the capacitor and create a current draw from it? The simple
circuit below accomplishes this feat.
The buffers B1 and B2 essentially transfer the capacitor voltage to a resistor R. The upshot being a current created through the R leg that looks a lot like the capacitor voltage, and in turn  as we observed before  looks like the current through an RL circuit! Buff1 also prevents Cs, Rs from drawing the current from node vs, upsetting the intended behavior.
How do we design this simulator? To get the same wave shape versus time for R, L leg, we choose the Rs, Cs leg to have the same time constant as the RL circuit: L / R = Rs * Cs. So choosing Cs, you solve for Rs to get
Rs = L / (R * Cs)
To get the same current draw simply set R to the resistance of the RL leg. For example, to simulate a R = 1100 and L = 0.33H, choose Cs=1nF and calculate
Rs = 0.33 / (1100 * 1nF)
= 300kR = 1100
GRAPHIC EQ CIRCUIT
Finally, after a long journey we are able to assemble the pieces of
graphic EQ together. We take the band cut/boost circuit from Part 3 and
replace the RL leg with our simulated RL!
The XOP2 devices acts as the Buff2 in the simplified diagram above. But wait  what happened to Buff1. To save components (space and money), a simple trick makes Buff1 unnecessary. By choosing CS and RS to have large impedances, the current flow in these components becomes negligible compared to R3. No buffer needed.
SPICE TEST
Let's simulate an EQ circuit designed for K=±10x
@ 800 Hz. Set RPOS Fully Counter
Clockwise (RPOS=1) for max attenuation. Run the file OP_L_SIM_BANDCUT_BOOST.CIR to
perform an AC Analysis on the circuit.
CIRCUIT INSIGHT Check the frequency response by plotting the input V(1) and output V(5). Does the output show gain or attenuation? Does it cut the signal at about 10x (or 20dB)? Is the attenuation wideband or narrowband? What is the center frequency?
Now, set RPOS Fully Clockwise (RPOS=99) for max boost and rerun the simulation. Does it hit the design target of a boost around 10x (or +20dB)?
Finally, set RPOS to the center position (RPOS=50) and rerun the
simulation. What do you think the response will be with the pot set dead
center between cut and boost positions?
FULL GRAPHIC EQ
Now that we've got an EQ circuit centered at 800Hz, how do you add 
say 5 other bands  to create a sixband Graphic EQ? The cool thing is you
don't have to duplicate the entire circuit! You'll need the following
Just one amplifier XOP1 and its gain resistors R1 and R4.
Then duplicate the pot plus LCR leg for each EQ band. Calculate new LCR values for each center frequency.
Lastly, all of the pots for each band are connected across the XOP1's + and  input terminals. How does this work? All of the band cut/boost filters work simultaneously, the op amp summing the action of each one.
SPICE FILE
Download the file or copy this netlist into a text file with the *.cir extention.
* op_bandcut_boost_L_sim.cir * VS 1 0 AC 1 SIN(0V 1V 800Hz) * * POTENTIOMETER POSITION (1=CCW, 50=MID, 99=CW) .PARAM RPOS=1 * R1 1 2 10K R2A 2 3 { RPOS/100*20K } R2B 3 4 { (1RPOS/100)*20K } R4 4 5 10k * OPAMP XOP1 2 4 5 OPAMP1 * C1 3 6 120NF * * SIMULATED LR CS 6 7 1NF RS 7 0 300K R3 6 8 1100 XOP2 7 8 8 OPAMP1 * * OPAMP MACRO MODEL, SINGLEPOLE * connections: noninverting input *  inverting input *   output *    .SUBCKT OPAMP1 1 2 6 * INPUT IMPEDANCE RIN 1 2 10MEG * GAIN BW PRODUCT = 1MHZ * DC GAIN (100K) AND POLE 1 (10HZ) EGAIN 3 0 1 2 100K RP1 3 4 1K CP1 4 0 15.915UF * OUTPUT BUFFER AND RESISTANCE EBUFFER 5 0 4 0 1 ROUT 5 6 10 .ENDS * * ANALYSIS ************************************* *.TRAN 1us 100us .ac dec 40 10 100k .PROBE .END
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